
\(M_z \in \mathcal {B}(H^2()\)  is an isometry and hence, left-invertible (see [1]}).


For \(-1< \alpha \le 0\) , the shift operator \(M_{z} \in \mathcal {B}(A_{\alpha }^2()\)  satisfies the operator identity \(\Vert M_z f + g\Vert ^2 \le 2 (\Vert f\Vert ^2 + \Vert M_z g\Vert ^2)\)  for \(f,g \in A_{\alpha }^2(\mathbb {D})\)  (see [Theorem 6.15, [2]}).


\(M_{z} \in \mathcal {B}(\mathcal {D}()\)  satisfies \(M_z^2 {M_z}^{*2} - 2 M_z M_z^* +I = 0\) , in other words, by Agler's convention \(M_z\)  is a 2-isometry and it follows that \(M_z\)  is left-invertible (see [3]}).

